Electric Potential Expression

[Screwdriver]

Homework Statement

Determine an expression for the electric potential at a point P a distance z away from the center of a thin uniformly charged rod on the line that bisects the rod.

Homework Equations

$$V=k\frac{q}{r}\hat{r}$$

The Attempt at a Solution

I've defined r to be the straight-line distance from any point, i, on the rod to P and y to be the vertical distance along the rod to i.

Determine the charge due to some point i:

$$V_i=k\frac{\Delta Q}{r}\hat{r}$$

I determined that:

$$\hat{r}=cos(\theta )=\frac{z}{\sqrt{y^2+z^2}}$$

So:

$$V_i=kz\frac{\Delta Q}{y^2+z^2}$$

Then with linear charge density:

$$Q=\lambda y\Rightarrow \Delta Q=\lambda \Delta y$$

$$V_i=kz\lambda \frac{\Delta y}{y^2+z^2}$$

Then sum those up and take the limit as the sum approaches infinity and as delta y approaches zero:

$$V_{net}=kz\lambda \int_{\frac{-L}{2}}^{\frac{L}{2}}\frac{d y}{y^2+z^2}=k\lambda arctan(\frac{y}{z})|_{\frac{-L}{2}}^{\frac{L}{2}}$$

$$V_{net}=2k\lambda arctan(\frac{L}{2z})$$

 

[anathema]

Hi there, great question! Your solution looks mostly correct, but there are a few things that could be improved upon.

Firstly, your expression for the unit vector, r^, is not quite correct. It should be r^ = (z/sqrt(y^2 + z^2), 0, y/sqrt(y^2 + z^2)). This can be derived using basic trigonometry and the Pythagorean theorem.

Also, when you are integrating to find the net potential, you should use the full length of the rod, L, instead of just half of it. This is because the charge on the entire rod contributes to the potential at point P, not just half of it.

Finally, when solving for the net potential, you should also include a factor of 2πε0 in your expression. This is a constant that is often left out in calculations, but it is important to include for accurate results.

With these adjustments, your final expression for the net potential should be:

V_net = 2πε0kλarctan(L/2z)

Overall, your approach and solution are correct, but just be sure to double check your calculations and include all necessary constants for a complete and accurate solution. Keep up the good work!